Answer to Question #215697 in Differential Geometry | Topology for Ahmed

Question

Find the envelope of the family of the curves

1) y=3px-p^3

Where p=parameter

2) y=mx-2am-am^3

Where m=parameter

Answer 1

1) The parametric equations of the envelope are defined by the system of equations

$$y=3px-p^3$$ $$0=3x-3p^2$$

$$p=\pm\sqrt{x}$$

$$y=3x\sqrt{x}-x\sqrt{x}=2x\sqrt{x}$$

Or

$$y=-3x\sqrt{x}+x\sqrt{x}=-2x\sqrt{x}$$

2) The parametric equations of the envelope are defined by the system of equations

$$y=mx-2am-am^3$$ $$0=x-2a-3am^2$$

$$m=\pm\sqrt{\dfrac{x-2a}{3a}}$$

$$y=x\sqrt{\dfrac{x-2a}{3a}}-2a\sqrt{\dfrac{x-2a}{3a}}-\dfrac{x-2a}{3}\sqrt{\dfrac{x-2a}{3a}}$$

$$=\dfrac{3x-6a-x+2a}{3}\sqrt{\dfrac{x-2a}{3a}}$$

$$=\dfrac{2(x-2a)^{3/2}}{3\sqrt{3a}}$$

Or

$$y=-x\sqrt{\dfrac{x-2a}{3a}}+2a\sqrt{\dfrac{x-2a}{3a}}+\dfrac{x-2a}{3}\sqrt{\dfrac{x-2a}{3a}}$$ $$=\dfrac{-3x+6a+x-2a}{3}\sqrt{\dfrac{x-2a}{3a}}$$

$$=-\dfrac{2(x-2a)^{3/2}}{3\sqrt{3a}}$$